1951 AHSME Problems/Problem 24

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Problem

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:

$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$

Solution

We have $2(2^n)=2^{n+1}$, and $2(2^{n+3})=2^{n+4}$. Thus, $\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}$. Factoring out a $2^{n+1}$ in the numerator, we get $\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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